# When is 1 watt not 1 watt?

Discussion in 'Electronics Homework Help' started by chopnhack, Sep 2, 2014.

1. ### chopnhack

1,573
352
Apr 28, 2014
You are given three resistors: two 4Ω resistors and one 6Ω resistor.
What is the value in Ohms ( Ω ) of the largest-valued resistor that can be fabricated by combining these three resistors?
14 - correct

What is the value in Ohms ( Ω ) of the smallest-valued resistor that can be fabricated by combining these three resistors?
1.5 - correct

Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the smallest-valued composite resistor dissipate before burning up?

Here is my stumbling block - if each resistor has the property of being able to dissipate 1 w, then 1w is the max it can dissipate before burning up. Am I looking at this too obtusely? Parallel network of resistors (4,4,6Ω) the smallest composite would be a combination of 4Ω+4Ω+6Ω yielding a composite resistor of 1.5Ω capable of dissipating 1W total being its in parallel, had it been in series I would say 3W. The computer says wrong. Can someone shed some light on this?

2. ### Gryd3

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Jun 25, 2014
In Parallel each resistor will dissipate energy.
The smallest element can dissipate 1W, now you need to figure out what the other elements will dissipate when the smallest element is dissipating 1W and add it to the sum.

(This is a basic reminder I know, but remember that the voltage across parallel components is the same)

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5,164
1,081
Dec 18, 2013
Ah just done that one. Look at the 4R and find out how much current can go through it to produce 1W this will be the same for the other 4R also. Work out the voltage required for this current and the use this to work out the current in the 6R and then the wattage. Then add them all together.

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4. ### Gryd3

4,098
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Jun 25, 2014
You guys in school together?

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1,573
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Apr 28, 2014
6. ### chopnhack

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Apr 28, 2014
Thanks to you both for clearing that up for me. I was able to get the correct mathematical figure of 2.66W but I am still not clear as to the why of it. I have been at it from 8:30AM-3PM and it has been a waste... I really need to revisit and get more clarity on nodes, kvl and kcl - specifically how to set up the equations. I now know as some have taught me here that I am trying to setup equations from known to unknown, but it still escapes me. I wish the course had more practice examples with explanations to learn from. Can't complain, the price is right 5,164
1,081
Dec 18, 2013
The way I looked at it was a bit different. I thought the 4Rs are going to be the limiting factor as they are going to draw more current and thus generate the 1W first. So I just worked from there, then your fixed with that voltage so the 6R just dissipates the x amount of power as a result of the fixed voltage.

8. ### Gryd3

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Jun 25, 2014
Algebra is your friend Just post when you hit another block. I'll start that course too chopnhack likes this.
9. ### chopnhack

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Apr 28, 2014
I thank you! It's a blessing to have you guys to help me from bashing my head into the concrete....  Algebra is good, I use some low level forms all the time, but I realize how deficient I have become in multiple variable equations and moving variables around... I need to review! Khan Academy will be my friend tonight, if I can finish my homework and start on the next weeks assignments. I can't imagine taking more than one class at a time, college was too many years ago!

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10. ### Laplace

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184
Apr 4, 2010
Did the course work ask the corresponding question:

Given that each individual resistor can dissipate up to 1 watt of power before burning up, how much total power in watts ( W ) can the largest-valued composite resistor dissipate before burning up?

What is the logic for finding that answer?

11. ### Gryd3

4,098
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Jun 25, 2014
Are you asking to test the OP, or is this curiosity?

1,252
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Apr 4, 2010
Both!

13. ### Gryd3

4,098
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Jun 25, 2014
Same type of math applies, but instead of the same voltage drop being across each resistor in parallel, the same current will be present in each resistor as they are in series.
You need to find the weakest link so to speak by finding the maximum amount of current that two 4R and a 6R can tolerate in series if each one is rated for 1W. Each resistor can be analyzed individually to determine which one will dissipate 1W at the lowest current. This lower current is then used to calculate the power dissipation of the remaining resistors since the current must be the same in all series connected components.

Formulas to help:
V = IR
P = IV

P = R * I^2 [Used to find power dissipated by the resistor once current is known]
I = Sqrt( P / R ) [Used to find max current given resistance and power rating]

5,164
1,081
Dec 18, 2013
The resistor are all in Parallel, in the last part of this question.

15. ### Gryd3

4,098
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Jun 25, 2014
The largest composite resistor value would require the resistors to be in series.

Sorry if I caused any confusion for you.

5,164
1,081
Dec 18, 2013
Oh sorry, just the question actually asked lowest, I wonder why Laplace said highest?

17. ### Laplace

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Apr 4, 2010
Because even if the coursework did not ask both questions, the OP should be able to answer both questions, and explain the difference.

5,164
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Dec 18, 2013
Oh ok

19. ### chopnhack

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Apr 28, 2014
All the coursework asked for was in post 1 - the highest resistance of 14Ω was achieved by series connection of the three resistors. As to how much power can the largest composite resistor dissipate without burning up:

Since it converts to a series circuit there is constant current at each element, voltage will drop across each element uniquely to each resistance thus balancing.
As Gyrd pointed out the math is similar in that you need to solve for current first.

Being that the resistors are commutative, I will use 6Ω as the first resistor that current will pass through.

P=IsquaredR

1W=6(I squared)
I=0.41A

P=0.41A*R
P=0.41*0.41(4)
P=0.168(4)
P=0.672W

1W+0.672W+0.672W
Total power dissipated by series = 2.344W

Now if these were resistors of the same values, then there would be an even split of current and thus maximum dissipation would be possible. Had I done the 4Ω resistor first, the 6Ω would be passed too much current and would probably burn open.

My next question is this - in the textbook they say that one of the tenets of circuit analysis is simplification - wherever there is a complex series or parallel circuit, try and reduce it to an equivalent resistance. I certainly like that idea, but if we employed that in this example, the answer would come out wrong? Can someone explain this contradiction to me, or have I simply missed something again?

above, series equivalent resistance would be sum of 4+4+6 = 14Ω. If we wanted to dissipate 1W over this.... ah... I think I get it, its a series equivalent, but its still three discrete items which you can't lump dissipate 1 watt over the three, they have to be individually evaluated for power dissipation, especially since they are of different values!!! 20. ### Gryd3

4,098
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Jun 25, 2014
Sounds good to me! You'd end up with a 'new' component capable of dissipating 2.34W with a resistance of 14Ω You could take these 3 items and plunk em in a more complex circuit and those 3 items together will still behave the same.

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